Problem: The first term of a given sequence is 1, and each successive term is the sum of all the previous terms of the sequence. What is the value of the first term which exceeds 5000?
Solution: We calculate the first several terms directly and find the sequence starts
\[ 1, 1, 2, 4, 8, 16, \ldots \] It appears the $n$th term is $2^{n-2}$ for $n\geq 2$.  Since $2^{12}=4096$, the first power of 2 that exceeds 5000 is $2^{13}=\boxed{8192}$.

Let's prove by induction that the $n$th term of the sequence is $2^{n-2}$ for all integers $n\geq 2$.  The base case $n=2$ holds since the second term of the sequence is the sum of all the terms before it, which is just 1.  For the induction step, let $n>2$ and suppose that the $(n-1)$st term is $2^{n-1-2}=2^{n-3}$.  Then the sum of the first $n-2$ terms of the sequence is $2^{n-3}$, since the $(n-1)$st term is equal to the sum of the first $n-2$ terms.  So the $n$th term, which is defined to be the sum of the first $n-1$ terms, is \[\underbrace{2^{n-3}}_{\text{sum of first }n-2\text{ terms}}+\underbrace{2^{n-3}}_{(n-1)\text{st term}}=2\cdot2^{n-3}=2^{n-2}.\]   This completes the induction step, so the statement is proved for all $n\geq 2$.